3.641 \(\int \frac {x}{(a+b \sqrt {c+d x})^2} \, dx\)

Optimal. Leaf size=95 \[ \frac {2 a \left (a^2-b^2 c\right )}{b^4 d^2 \left (a+b \sqrt {c+d x}\right )}+\frac {2 \left (3 a^2-b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}-\frac {4 a \sqrt {c+d x}}{b^3 d^2}+\frac {x}{b^2 d} \]

[Out]

x/b^2/d+2*(-b^2*c+3*a^2)*ln(a+b*(d*x+c)^(1/2))/b^4/d^2-4*a*(d*x+c)^(1/2)/b^3/d^2+2*a*(-b^2*c+a^2)/b^4/d^2/(a+b
*(d*x+c)^(1/2))

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Rubi [A]  time = 0.09, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {371, 1398, 772} \[ \frac {2 a \left (a^2-b^2 c\right )}{b^4 d^2 \left (a+b \sqrt {c+d x}\right )}+\frac {2 \left (3 a^2-b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}-\frac {4 a \sqrt {c+d x}}{b^3 d^2}+\frac {x}{b^2 d} \]

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Sqrt[c + d*x])^2,x]

[Out]

x/(b^2*d) - (4*a*Sqrt[c + d*x])/(b^3*d^2) + (2*a*(a^2 - b^2*c))/(b^4*d^2*(a + b*Sqrt[c + d*x])) + (2*(3*a^2 -
b^2*c)*Log[a + b*Sqrt[c + d*x]])/(b^4*d^2)

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,
x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]
] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rule 1398

Int[((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> With[{g = Denominator[n]}, D
ist[g, Subst[Int[x^(g - 1)*(d + e*x^(g*n))^q*(a + c*x^(2*g*n))^p, x], x, x^(1/g)], x]] /; FreeQ[{a, c, d, e, p
, q}, x] && EqQ[n2, 2*n] && FractionQ[n]

Rubi steps

\begin {align*} \int \frac {x}{\left (a+b \sqrt {c+d x}\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {-c+x}{\left (a+b \sqrt {x}\right )^2} \, dx,x,c+d x\right )}{d^2}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {x \left (-c+x^2\right )}{(a+b x)^2} \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (-\frac {2 a}{b^3}+\frac {x}{b^2}+\frac {-a^3+a b^2 c}{b^3 (a+b x)^2}+\frac {3 a^2-b^2 c}{b^3 (a+b x)}\right ) \, dx,x,\sqrt {c+d x}\right )}{d^2}\\ &=\frac {x}{b^2 d}-\frac {4 a \sqrt {c+d x}}{b^3 d^2}+\frac {2 a \left (a^2-b^2 c\right )}{b^4 d^2 \left (a+b \sqrt {c+d x}\right )}+\frac {2 \left (3 a^2-b^2 c\right ) \log \left (a+b \sqrt {c+d x}\right )}{b^4 d^2}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 112, normalized size = 1.18 \[ \frac {2 a^3+2 \left (3 a^2-b^2 c\right ) \left (a+b \sqrt {c+d x}\right ) \log \left (a+b \sqrt {c+d x}\right )-4 a^2 b \sqrt {c+d x}-3 a b^2 (2 c+d x)+b^3 d x \sqrt {c+d x}}{b^4 d^2 \left (a+b \sqrt {c+d x}\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Sqrt[c + d*x])^2,x]

[Out]

(2*a^3 - 4*a^2*b*Sqrt[c + d*x] + b^3*d*x*Sqrt[c + d*x] - 3*a*b^2*(2*c + d*x) + 2*(3*a^2 - b^2*c)*(a + b*Sqrt[c
+ d*x])*Log[a + b*Sqrt[c + d*x]])/(b^4*d^2*(a + b*Sqrt[c + d*x]))

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fricas [A]  time = 0.47, size = 163, normalized size = 1.72 \[ \frac {b^{4} d^{2} x^{2} + b^{4} c^{2} + a^{2} b^{2} c - 2 \, a^{4} + {\left (2 \, b^{4} c - a^{2} b^{2}\right )} d x - 2 \, {\left (b^{4} c^{2} - 4 \, a^{2} b^{2} c + 3 \, a^{4} + {\left (b^{4} c - 3 \, a^{2} b^{2}\right )} d x\right )} \log \left (\sqrt {d x + c} b + a\right ) - 2 \, {\left (2 \, a b^{3} d x + 3 \, a b^{3} c - 3 \, a^{3} b\right )} \sqrt {d x + c}}{b^{6} d^{3} x + {\left (b^{6} c - a^{2} b^{4}\right )} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^(1/2))^2,x, algorithm="fricas")

[Out]

(b^4*d^2*x^2 + b^4*c^2 + a^2*b^2*c - 2*a^4 + (2*b^4*c - a^2*b^2)*d*x - 2*(b^4*c^2 - 4*a^2*b^2*c + 3*a^4 + (b^4
*c - 3*a^2*b^2)*d*x)*log(sqrt(d*x + c)*b + a) - 2*(2*a*b^3*d*x + 3*a*b^3*c - 3*a^3*b)*sqrt(d*x + c))/(b^6*d^3*
x + (b^6*c - a^2*b^4)*d^2)

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giac [A]  time = 0.41, size = 102, normalized size = 1.07 \[ -\frac {\frac {2 \, {\left (b^{2} c - 3 \, a^{2}\right )} \log \left ({\left | \sqrt {d x + c} b + a \right |}\right )}{b^{4} d} - \frac {{\left (d x + c\right )} b^{2} d - 4 \, \sqrt {d x + c} a b d}{b^{4} d^{2}} + \frac {2 \, {\left (a b^{2} c - a^{3}\right )}}{{\left (\sqrt {d x + c} b + a\right )} b^{4} d}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^(1/2))^2,x, algorithm="giac")

[Out]

-(2*(b^2*c - 3*a^2)*log(abs(sqrt(d*x + c)*b + a))/(b^4*d) - ((d*x + c)*b^2*d - 4*sqrt(d*x + c)*a*b*d)/(b^4*d^2
) + 2*(a*b^2*c - a^3)/((sqrt(d*x + c)*b + a)*b^4*d))/d

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maple [A]  time = 0.01, size = 125, normalized size = 1.32 \[ -\frac {2 a c}{\left (a +\sqrt {d x +c}\, b \right ) b^{2} d^{2}}-\frac {2 c \ln \left (a +\sqrt {d x +c}\, b \right )}{b^{2} d^{2}}+\frac {x}{b^{2} d}+\frac {2 a^{3}}{\left (a +\sqrt {d x +c}\, b \right ) b^{4} d^{2}}+\frac {6 a^{2} \ln \left (a +\sqrt {d x +c}\, b \right )}{b^{4} d^{2}}+\frac {c}{b^{2} d^{2}}-\frac {4 \sqrt {d x +c}\, a}{b^{3} d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+(d*x+c)^(1/2)*b)^2,x)

[Out]

x/b^2/d+1/d^2/b^2*c-4*a*(d*x+c)^(1/2)/b^3/d^2-2/d^2*a/b^2/(a+(d*x+c)^(1/2)*b)*c+2/d^2*a^3/b^4/(a+(d*x+c)^(1/2)
*b)-2/d^2/b^2*ln(a+(d*x+c)^(1/2)*b)*c+6/d^2/b^4*ln(a+(d*x+c)^(1/2)*b)*a^2

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maxima [A]  time = 1.05, size = 90, normalized size = 0.95 \[ -\frac {\frac {2 \, {\left (a b^{2} c - a^{3}\right )}}{\sqrt {d x + c} b^{5} + a b^{4}} - \frac {{\left (d x + c\right )} b - 4 \, \sqrt {d x + c} a}{b^{3}} + \frac {2 \, {\left (b^{2} c - 3 \, a^{2}\right )} \log \left (\sqrt {d x + c} b + a\right )}{b^{4}}}{d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)^(1/2))^2,x, algorithm="maxima")

[Out]

-(2*(a*b^2*c - a^3)/(sqrt(d*x + c)*b^5 + a*b^4) - ((d*x + c)*b - 4*sqrt(d*x + c)*a)/b^3 + 2*(b^2*c - 3*a^2)*lo
g(sqrt(d*x + c)*b + a)/b^4)/d^2

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mupad [B]  time = 0.06, size = 98, normalized size = 1.03 \[ \frac {x}{b^2\,d}+\frac {2\,\left (a^3-a\,b^2\,c\right )}{b\,\left (b^4\,d^2\,\sqrt {c+d\,x}+a\,b^3\,d^2\right )}-\frac {\ln \left (a+b\,\sqrt {c+d\,x}\right )\,\left (2\,b^2\,c-6\,a^2\right )}{b^4\,d^2}-\frac {4\,a\,\sqrt {c+d\,x}}{b^3\,d^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*(c + d*x)^(1/2))^2,x)

[Out]

x/(b^2*d) + (2*(a^3 - a*b^2*c))/(b*(b^4*d^2*(c + d*x)^(1/2) + a*b^3*d^2)) - (log(a + b*(c + d*x)^(1/2))*(2*b^2
*c - 6*a^2))/(b^4*d^2) - (4*a*(c + d*x)^(1/2))/(b^3*d^2)

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sympy [A]  time = 42.38, size = 131, normalized size = 1.38 \[ \begin {cases} \frac {2 \left (- \frac {a \left (a^{2} - b^{2} c\right ) \left (\begin {cases} \frac {\sqrt {c + d x}}{a^{2}} & \text {for}\: b = 0 \\- \frac {1}{b \left (a + b \sqrt {c + d x}\right )} & \text {otherwise} \end {cases}\right )}{b^{3} d} - \frac {2 a \sqrt {c + d x}}{b^{3} d} + \frac {c + d x}{2 b^{2} d} + \frac {\left (3 a^{2} - b^{2} c\right ) \left (\begin {cases} \frac {\sqrt {c + d x}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b \sqrt {c + d x} \right )}}{b} & \text {otherwise} \end {cases}\right )}{b^{3} d}\right )}{d} & \text {for}\: d \neq 0 \\\frac {x^{2}}{2 \left (a + b \sqrt {c}\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*(d*x+c)**(1/2))**2,x)

[Out]

Piecewise((2*(-a*(a**2 - b**2*c)*Piecewise((sqrt(c + d*x)/a**2, Eq(b, 0)), (-1/(b*(a + b*sqrt(c + d*x))), True
))/(b**3*d) - 2*a*sqrt(c + d*x)/(b**3*d) + (c + d*x)/(2*b**2*d) + (3*a**2 - b**2*c)*Piecewise((sqrt(c + d*x)/a
, Eq(b, 0)), (log(a + b*sqrt(c + d*x))/b, True))/(b**3*d))/d, Ne(d, 0)), (x**2/(2*(a + b*sqrt(c))**2), True))

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